this one is tricky and freaky. But easy. Find all its

Question

this one is tricky and freaky. But easy.
Find all its solutions

(2C)!–|C|=C

stolyar · Answer

Alas, you are wrong again. How did you manage to be admitted by doing so many mistakes? What about 1/2?

For those who still doubt I can post a solution.

kpadma · Answer

How does one know the one got all the possible solutions to such equations? Are there any rules ?

stolyar · Answer

(2C)!=C+|C|

Since C is inside a factorial, C >=0, so open up a modul.

(2C)!=C+C=2C
Let's 2C=A, we have A!=A; thus A=1 or 2

2C=1, thus C=1/2
2C=2, thus C=1

no other roots!

simple, isn't it?

AkamaiBrah · Answer

Hi!

Yes. There are rules. 

I teach, IMHO, a very good and systematic method to solve ANY type of absolute value problem. Learn this method because it works EVERY TIME.

1) first determine the "critical points" of the equation. The critical points are those point at which a complete expression inside one of the absolute value sign changes from positive to negative or vice-versa. 

2) Rewrite the equations in the intervals between +infinity, -infinity, and all of the critical points WITHOUT absolute value signs.
3) Solve for each interval. IF the solution(s) are not compatible with the interval you are in, then that solution is not valid. 

Repeat steps 2 and 3 for each interval.


I know it's a little confusing, but let me give you a few examples:

Ex 1: (Easy example you can do in your head)

Solve: |x - 1| = 4

Step 1: 
The critical point here is x=1 because at this point, the expression x - 1 changes sign. 

Steps 2 and 3: 
The important intervals are x 1. (It is okay to include the endpoint in the intervals). 
When x 1, the expression x - 1 is always positive, so the absval signs will have no effect on the expression. So for this interval, lets rewrite it as:
x - 1 = 4, and solving, we get x = 5. The solution is in the interval (x > 1) so it is also a valid solution, thus the solution set is x = {-3,5}.


Example 2: (You can do this one in your head only if you are Russian or Chinese (they learn Calculus before puberty), but it's real easy to get confused). 

Solve: |x тАУ 2| - |x тАУ 3| = |x тАУ 5|?

Step 1:
Recall, the critical points are -inf, +inf, and any finite numbers that causes any complete expression within an absolute value pair to change signs from negative to positive or vice-versa. For example, as x goes from less than 2 to greater than 2, тАЬx тАУ 2тАЭ тАУ the expression inside the first pair of absolute value signs тАУ goes from negative to positive. Hence, 2 is one of the critical points. By the same logic, we can deduce that the other two critical points are 3 and 5. 

Steps 2 and 3:
Since the finite critical point is 2, we will choose our first interval of interest as -inf = 0. Rewriting the equation so that it would always be true in this interval, we get 

 (2C)! = C + C or (2C!) = 2C. 

Now divide both sides by 2C, we get: 

 (2C - 1)! = 1. 

Well, (2C - 1)! will equal 1 only when (2C - 1) = 0 or 1, hence C must equal either 1/2 or 1. Hence, the solution is C = {1/2, 1}.

Note: it is possible to write equations that make this method a little tedious (i've seen a few of Stoylar's equation where there are nested abs val signs such as: ||X^2| - 2X| = ||X + 4| - (|X| - 3)| -- yes, he can be a sadist.)

Two relevant comments: 1) you can still use this method -- it's just a little trickier to determine what the critical points are; and 2) you will NEVER get a problem like this on the GMAT, so don't worry, be happy. 

Hope this adds a nice tool to your toolbox!

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